Hi
Hope you can help me with my problem. Hope you understand my English :)
A phone company asks 50 people about their waiting time, when they call the customer service.
The sample mean is 8.4 minutes and the standard deviation is 7.8 min.
We also know that 10 out of the 50 people are waiting for longer than 15 min.
So I know that ´x = 8.4 and s = 7.8.
Question 1:
Determine the 95% confidence interval estimate of the mean waiting time
I solve this question with the t interval formula.

But I am lost at the next questionQuestion 2:I am supposed to make a hypotheses test for the proportion. Can the company clam that under ¼ of the customers are waiting for UNDER 15min??
I am lost??
How can I do that?? Well I am not quite sure how to define Ho and H1?
Hope you can help me :)
And thank you all for sharing your knowledge! :)

Dec 7th 2012, 05:34 PM

abender

Re: proportion

Quote:

Originally Posted by Jonas8000

Hi
Hope you can help me with my problem. Hope you understand my English :)
A phone company asks 50 people about their waiting time, when they call the customer service.
The sample mean is 8.4 minutes and the standard deviation is 7.8 min.
We also know that 10 out of the 50 people are waiting for longer than 15 min.
So I know that ´x = 8.4 and s = 7.8.
Question 1:
Determine the 95% confidence interval estimate of the mean waiting time
I solve this question with the t interval formula.

But I am lost at the next questionQuestion 2:I am supposed to make a hypotheses test for the proportion. Can the company clam that under ¼ of the customers are waiting for UNDER 15min??
I am lost??
How can I do that?? Well I am not quite sure how to define Ho and H1?
Hope you can help me :)
And thank you all for sharing your knowledge! :)

$\displaystyle \fbox{Question 1:}$
Since the sample size, $\displaystyle n=50$, is sufficiently large (greater than 30), you can go ahead and use Z instead of t with 49 df. If this concept does not make sense to you, familiarize yourself with the Central Limit Theorem. Note that the table values for Z and for t with 49 df will be very close to one another.
That said, your confidence interval takes on the form $\displaystyle \bar{X}\pm\frac{S}{\sqrt{n}}Z_{\alpha/2}$. Since you know $\displaystyle \bar{X}$, $\displaystyle S$, $\displaystyle n$, and $\displaystyle \alpha=0.5 \implies \tfrac{\alpha}{2}=0.025$, computing your desired confidence interval is straightforward.

Dec 7th 2012, 06:53 PM

abender

Re: proportion

Quote:

Originally Posted by Jonas8000

Hi
A phone company asks 50 people about their waiting time, when they call the customer service.
The sample mean is 8.4 minutes and the standard deviation is 7.8 min.
We also know that 10 out of the 50 people are waiting for longer than 15 min.
So I know that ´x = 8.4 and s = 7.8.Question 2:I am supposed to make a hypotheses test for the proportion. Can the company clam that under ¼ of the customers are waiting for UNDER 15min??
I am lost??
How can I do that?? Well I am not quite sure how to define Ho and H1?

Did you write this question down correctly? Your sample suggests 40/50=0.8 of customers wait for fewer than 15 minutes, and you want to test the claim that under 0.25 of customers wait under 15 minutes? The rejection region for this test is $\displaystyle Z<-Z_{\alpha}\approx -1.645$, $\displaystyle Z$ ends up very large (like, 9). So, clearly, we fail to reject $\displaystyle H_0$, and say there is not statistical evidence for the claim at $\displaystyle \alpha=0.05$.

This doesn't sound like a question that would be asked. I would imagine the question asks to test the claim that under 0.25 of customers wait OVER 15 minutes. Can you confirm this for me, please?